This is the Grade 10 Mathematics Self-Learning Module for Quarter 2, Module 8: “Writing Equation of a Circle and Determining the Center and Radius of a Circle.” It was developed by DepEd Cordillera Administrative Region (First Edition, 2020) for learners studying at home under the Alternative Delivery Mode. After working through it, learners can write the equation of a circle from its center and radius, and find the center and radius when they are given the equation.
On this page 8 sections
What learners will learn
By the end of the module, learners should be able to:
- illustrate the center-radius form of the equation of a circle, and
- determine the center and radius of a circle given its equation, and vice versa.
The module explains that this topic is essential in coordinate geometry, which is used in fields such as mapping, aeronautics, and navigation.
Review: circles and the distance formula
The module starts by recalling two ideas from earlier lessons:
- Circle: the set of all points (x, y) on a plane that are the same distance from a fixed point, the center. That distance is the radius, r.
- Distance formula: the distance between A(x₁, y₁) and B(x₂, y₂) is d = √((x₂ − x₁)² + (y₂ − y₁)²).
In Activity 1, learners use the distance formula to find the radius of two graphed circles. One has center (0, 0) and passes through (2, 2), giving a radius of 2√2. The other has center (1, 2) and passes through (−2, 1), giving a radius of √10.
The two forms of the equation of a circle
Center-radius form (standard form)
The module uses the distance formula to build the equation. It finds the distance between the center and any point P(x, y) on the circle.
| Center | Equation | Example |
|---|---|---|
| At the origin (0, 0) | x² + y² = r² | x² + y² = 9 |
| At any point (h, k) | (x − h)² + (y − k)² = r² | (x − 3)² + (y + 1)² = 4 |
General form
Expanding the standard form gives the general form: x² + y² + Cx + Dy + E = 0, where C, D, and E are real numbers. Examples from the module include x² + y² − 6x − 10y + 18 = 0 and x² + y² − 8x = 0.
The module also gives these conditions:
- The radius must be positive.
- The coefficients of x² and y² are usually 1. If they are not 1, they must be equal to each other and not zero. An example is 4x² + 4y² − 16x + 24y − 36 = 0.
Worked examples in the module
The module solves six examples step by step:
- x² + y² = 25: the center is (0, 0) and the radius is 5, the square root of 25.
- x² + (y − 4)² = 16: rewrite as (x − 0)² + (y − 4)² = 4². The center is (0, 4) and the radius is 4.
- x² + y² − 6x − 10y + 18 = 0: group the x terms and the y terms, move 18 to the other side, and complete the squares by adding (6/2)² = 9 and (10/2)² = 25. This gives (x − 3)² + (y − 5)² = 16, so the center is (3, 5) and the radius is 4.
- 4x² + 4y² − 16x + 24y − 36 = 0: first divide every term by 4, then complete the squares. The result is (x − 2)² + (y + 3)² = 22, so the center is (2, −3) and the radius is √22.
- Center (6, 7), passing through (2, 5): use the distance formula to get r = √20 = 2√5. The equation is (x − 6)² + (y − 7)² = 20.
- Center (−5, −4), radius 3, in general form: write (x + 5)² + (y + 4)² = 9, expand, and simplify to x² + y² + 10x + 8y + 32 = 0.
A quick tip from these examples: the signs flip. In (x − 3)², h is 3, and in (y + 4)², k is −4.
Activities and assessment
- What I Know (pretest): 10 multiple-choice items on standard and general forms, centers, and radii.
- Activity 2: Pair them Up: match 15 sets of center and radius with their equations, then use the letters to decode a hidden message.
- What I Have Learned: fill in six statements that summarize the lesson.
- What I Can Do: a word problem about Mang Cardo’s carabao, which runs in a circle around its peg. Learners find the length of the rope from the equation x² + y² − 6x + 8y − 119 = 0, using a scale of 1 meter to 2 units.
- Assessment (post-test): 10 multiple-choice items.
- Additional Activity: match five graphed circles with their equations.
How to use this module
- Learners: Answer the pretest first to see what you already know. Then work through the parts in order. Write your answers on a separate sheet of paper, not on the module. Check your work honestly with the answer key at the end, and ask your teacher or facilitator if a task is unclear.
- Parents and facilitators: Help your child follow the parts in order. Remind them to finish the post-test before looking at the answer key.
- Teachers: The answer key covers the pretest, Activity 2, the word problem, the post-test, and the additional activity. Check it before you hand it out. Worked out from the module’s own method, pretest item 6 gives center (2, −5), which is choice C, but the key lists B. Activity 2 item 15 is O, which is needed to decode the message “OH LEARNING MATH IS REALLY FUN!”, but the key lists N.
Frequently asked questions
What grade and quarter is this module for?
It is the Grade 10 Mathematics Self-Learning Module for Quarter 2, Module 8. It was developed by DepEd Cordillera Administrative Region.
What is the center-radius form of the equation of a circle?
It is x² + y² = r² for a circle centered at the origin. For a circle centered at (h, k), it is (x − h)² + (y − k)² = r².
How do you find the center and radius from the general form?
Group the x terms and the y terms, move the constant to the other side, and complete the square for each variable. This rewrites the equation as (x − h)² + (y − k)² = r², so you can read off the center and the radius. If the coefficients of x² and y² are not 1, divide the whole equation by that coefficient first.
Does the module have an answer key?
Yes. The answer key covers the pretest, Activity 2, the word problem, the post-test, and the additional activity. A few pretest and Activity 2 answers appear to be misprinted, so check them against the worked method.
How long is the carabao’s rope in the What I Can Do problem?
According to the answer key, the rope measures 6 meters. Completing the square gives a radius of 12 units, which is 6 meters at a scale of 1 meter to 2 units.
Full text of the module
The text below is a transcription of the PDF above, as printed.
Mathematics
Quarter 2 – Module 8: Writing Equation of a Circle and Determining the Center and Radius of a Circle
10
CO_Q2_Mathematics 10_ Module 8
Mathematics – Grade 10 Alternative Delivery Mode Quarter 2 – Module 8: Writing Equation of a Circle and Determining the Center and Radius of a Circle First Edition, 2020
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Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them.
Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio
Development Team of the Module
Writers’ Name: Esther B. Guitobon and Cerion T. Camhit Editor’s Name: Melchor B. Ticag Reviewer’s Name: Bryan A. Hidalgo Management Team: May B. Eclar Benedicta B. Gamatero Carmel F. Meris Ethielyn E. Taqued Edgar H. Madlaing Marciana M. Aydinan Lydia I. Belingon
Printed in the Philippines by:
Department of Education – Cordillera Administrative Region
Office Address: Wangal, La Trinidad, Benguet Telefax: (074) 422-4074 E-mail Address: car@deped.gov.ph
10
Mathematics Quarter 2 – Module 8: Writing Equation of a Circle and Determining the Center and Radius of a Circle
Introductory Message
This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.
Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.
Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.
In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning.
Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.
If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.
Thank you.
What I Need to Know
This module contains activities, discussions and practice exercises on how to determine the center and radius of a circle given its equation and vice versa. This particular topic is essential in Coordinate Geometry. Understanding coordinate geometry has important applications like mapping in aeronautics and navigation, finding the distance and midpoints between points and many more in other different fields. Thus, it is important for you to know how to determine the center and radius of circles given its equation and vice versa.
At the end of this module, you will be able to achieve the following objectives:
- illustrate the center-radius form of the equation of a circle, and
- determine the center and radius of a circle given its equation and vice versa.
What I Know
DIRECTION: Let us determine how much you already know about the equation of a circle. Read and understand each item, then choose the letter of your answer and write it on your answer sheet.
Which equation represents an equation of a circle in the standard form?
- A) x + y = 7
- B) x + y² = 8
- C) x² + 4x + 2 = 0
- D) x² + y² = 25
Which of the following represents an equation of a circle in the general form?
- A) x² + 2x − 4y + 5 = 0
- B) y² − 3x + y + 5 = 0
- C) x² + y² + 4x − 4y − 28 = 0
- D) none of these
A circle defined by x² + y² = 9 has a graph whose center is at ___________.
- A) origin
- B) quadrant I
- C) quadrant II
- D) quadrant III
A circle with an equation defined by (x − 5)² + (y + 3)² = 9 has a graph whose center is located at ___________.
- A) quadrant I
- B) quadrant II
- C) quadrant III
- D) quadrant IV
What is the center of the circle defined by (x + 5)² + (y + 3)² = 36?
- A) (5,3)
- B) (−5,3)
- C) (−5, −3)
- D) (5, −3)
What is the center of the circle x² + y² − 4x + 10y + 13 = 0?
- A) (2,5)
- B) (−2,5)
- C) (2, −5)
- D) (−2, −5)
What is the radius of the circle x² + y² − 4x + 10y + 13 = 0?
- A) 3
- B) 4
- C) 5
- D) 6
What is the equation of the circle whose center is at (0,5) and has a radius of 5?
- A) x² + (y − 5)² = 25
- B) x² + y² = 25
- C) x² + (y − 5)² = 5
- D) (x − 5)² + y² = 25
What is the general form of the equation of a circle whose center is at (3,0) and the radius is 4?
- A) x² + y² − 4x + 10y + 13 = 0
- B) x² + y² − 10x + 10y + 16 = 0
- C) x² + y² − 6x − 7 = 0
- D) x² + y² − 6x + 7 = 0
Given by the equation of the circle, x² + y² + 4x − 4y − 28 = 0, what is its center-radius form?
- A) (x + 2)² + (y − 2)² = 36
- B) (x − 2)² + (y − 2)² = 36
- C) (x − 2)² + (y + 2)² = 36
- D) x² + y² = 36
What’s In
Your previous lessons on circles and the use of distance formula are both significantly relevant skills for you to understand the next lesson. Let’s have a short recall of these topics.
- CIRCLE
A circle is the set of all points, (x, y), on a plane having the same distance from a fixed point called the center of the circle. The distance between the center of the circle and any point on the circle is called the radius, r, of the circle.
- DISTANCE FORMULA
Supposing the coordinates of point A is (x₁, y₁) and point B is (x₂, y₂), then the distance, d, between point A and point B is
d = √((x₂ − x₁)² + (y₂ − y₁)²)
What’s New
Before you proceed to the lesson proper, it is important that you know how to use the distance formula in finding the measure of the radius of a circle. The following activity will also help you understand the lesson.
Activity 1: Radius of a Circle
Direction: Given the graph of each circle, find the length of the radius r using the distance formula.
| 1) | 2) | |
|---|---|---|
| Figure | [chart: coordinate graph; circle with center (0,0), radius r drawn to point (2, 2)] Labels: (2, 2), r, (0,0) | [chart: coordinate graph; circle with center (1, 2), radius r drawn to point (−2, 1)] Labels: (1, 2), r, (−2, 1) |
| Solution | ||
| Answer | Answer: ________________________ | Answer: ________________________ |
You did a good job in performing the given tasks! Now, it’s your time to check your own work. If you answered both items correctly, you may proceed to the next part of this module. If not, please try again.
check your answers
- 2√2 2) √10 [transcriber’s note: the answers are printed upside down, rotated 180 degrees, beside the “check your answers” label]
What Is It
You’ve learned already how to find the measure / length of the radius of a circle using the distance formula when given the coordinates of the center and a point on the circle. This time you will be learning about the different forms of the equation of a circle.
A. Circle with center at the origin.
Given the circle with center at (0,0). To draw its radius (r), connect the center to any point on the circle like point P(x, y). Then using the distance formula, we can find the equation of the circle.
We find the distance between the center(0,0) and P(x, y), which is the length of the radius(r) of the circle.
[chart: coordinate graph; circle centered at c(0,0) with radius r drawn to point P(x,y)]
From the distance formula, let P₁ (0,0), P₂ (x, y), and r be the distance.
d = √((x₂ − x₁)² + (y₂ − y₁)²) r = √((x − 0)² + (y − 0)²) Substitute P₁, P₂ and r in the formula. r = √(x² + y²) Simplify. r² = x² + y² Square both sides of the equation.
Thus, the equation of a circle with center at the origin is x² + y² = r².
B. Circle with center NOT at the origin.
Notice that the center of the circle is not on the origin but rather at point (h, k). At point P(x, y) on the circle, the radius r is drawn connecting the two points.
We find the distance between the center (h, k) and P(x, y), which is the length of the radius (r) of the circle.
[chart: coordinate graph; circle centered at c(h,k) with radius r drawn to point P(X,Y)]
From the distance formula, let 𝑃₁ (ℎ, 𝑘), 𝑃₂ (𝑥, 𝑦), and 𝑟 be the distance.
𝑑 = √((𝑥₂ − 𝑥₁)² + (𝑦₂ − 𝑦₁)²)
𝑟 = √((𝑥 − ℎ)² + (𝑦 − 𝑘)²) Substitute 𝑃₁ and 𝑃₂ in the formula.
𝑟² = (𝑥 − ℎ)² + (𝑦 − 𝑘)² Square both sides of the equation.
Thus, the equation of a circle with center not on the origin at any point (ℎ, 𝑘) is (𝒙 − 𝒉)² + (𝒚 − 𝒌)² = 𝒓².
Center-Radius Form (Standard Form) of the Equation of a Circle
Let 𝑟 be the radius of a circle, the equation of a circle in center-radius form is:
(a) 𝒙² + 𝒚² = 𝒓² with center at the origin (0,0)
(b) (𝒙 − 𝒉)² + (𝒚 − 𝒌)² = 𝒓² with center not on the origin at (ℎ, 𝑘)
Examples:
- 𝑥² + 𝑦² = 9
- (𝑥 − 3)² + (𝑦 + 1)² = 4
Expanding the standard form, we will arrive at the general form of the equation of a circle.
Expand: (x − h)² + (y − k)² = r²
Solution: x² − 2hx + h² + y² − 2ky + k² − r² = 0
x² + y² + (−2h)x + (−2k)y + h² + k² − r² = 0
*let C = −2h, D = −2k and E = h² + k² − r²
x² + y² + Cx + Dy + E = 0
General Form of the Equation of a Circle
Thus, the equation of a circle in general form is 𝐱² + 𝐲² + 𝐂𝐱 + 𝐃𝐲 + 𝐄 = 𝟎 where 𝐶, 𝐷 and 𝐸 are real numbers.
Examples:
- 𝑥² + 𝑦² − 6𝑥 − 10𝑦 + 18 = 0
- 𝑥² + 𝑦² − 8𝑥 = 0
- 𝑥² + 𝑦² + 2𝑦 − 3 = 0
- 4𝑥² + 4𝑦² − 16𝑥 + 24𝑦 − 36 = 0
Note: The numerical coefficients of 𝑥² and 𝑦² are both equal to 1, and the value of the radius (𝑟) is positive. Moreover, the numerical coefficients of 𝑥² and 𝑦² can be possibly not equal to 1 but they must be equal and cannot be equal to zero. See example 4 above.
Thus, the equation of a circle in general form is 𝐱² + 𝐲² + 𝐂𝐱 + 𝐃𝐲 + 𝐄 = 𝟎 where 𝐶, 𝐷 and 𝐸 are real numbers.
After learning the equations of a circle in relation to its center and radius. This time, you will learn how to determine the center and radius of a circle given the equation, and to write the equation of a circle given the center and the radius.
Example 1: Find the center and radius of the circle 𝑥² + 𝑦² = 25.
Solution: The given equation is in the form 𝑥² + 𝑦² = 𝑟². This means that the center of the circle is at the origin. To find the radius, get the square root of 25, which is 5.
Note: Since 𝑥² + 𝑦² = 25 is in the form 𝑥² + 𝑦² = 𝑟², then 𝑟² = 25. By extracting the square roots, 𝑟 = 5.
Therefore, the center of the circle is at (0, 0) and the radius is 5 units.
Example 2: Find the center and radius of the circle 𝑥² + (𝑦 − 4)² = 16.
Solution: The given equation is in the form (𝑥 − ℎ)² + (𝑦 − 𝑘)² = 𝑟². This means that the center of the circle is at (ℎ, 𝑘).
𝑥² + (𝑦 − 4)² = 16 Given.
(𝑥 − 0)² + (𝑦 − 4)² = 4² Write as (𝑥 − ℎ)² + (𝑦 − 𝑘)² = 𝑟².
ℎ = 0, 𝑘 = 4, 𝑟 = 4 Identify the values of ℎ, 𝑘, and 𝑟.
Therefore, the center of the circle is at (0, 4) and the radius is 4 units.
Example 3: Find the center and radius of the circle whose equation is represented by 𝑥² + 𝑦² − 6𝑥 − 10𝑦 + 18 = 0.
Solution: Since the equation is written in general form, we need to transform the equation to its standard form to determine its center and radius.
𝑥² + 𝑦² − 6𝑥 − 10𝑦 + 18 = 0 Given.
(𝑥² − 6𝑥) + (𝑦² − 10𝑦) + 18 = 0 Group terms with the same variable.
(𝑥² − 6𝑥) + (𝑦² − 10𝑦) = −18 Apply Addition Property of Equality.
(𝑥² − 6𝑥 + __) + (𝑦² − 10𝑦 + __) = −18 + __ + __ Apply completing the squares.
(𝑥² − 6𝑥 + 9) + (𝑦² − 10𝑦 + 25) = −18 + 9 + 25 (6/2)² = 9 (10/2)² = 25
(𝑥 − 3)² + (𝑦 − 5)² = 16 Factoring perfect square trinomial.
(𝑥 − 3)² + (𝑦 − 5)² = 4² Write as (𝑥 − ℎ)² + (𝑦 − 𝑘)² = 𝑟².
ℎ = 3, 𝑘 = 5, 𝑟 = 4 Identify the values of ℎ, 𝑘, and 𝑟.
Therefore, the center of the circle is at (3, 5) and the radius is 4 units.
Example 4: Find the center and radius of the circle whose equation is represented by 4𝑥² + 4𝑦² − 16𝑥 + 24𝑦 − 36 = 0.
Solution: Since the numerical coefficients of 𝑥² and 𝑦² are both not equal to 1, divide both sides of the equation by 4.
(4𝑥² + 4𝑦² − 16𝑥 + 24𝑦 − 36)/4 = 0/4 Divide each term by 4.
𝑥² + 𝑦² − 4𝑥 + 6𝑦 − 9 = 0
(𝑥² − 4𝑥) + (𝑦² + 6𝑦) − 9 = 0
(𝑥² − 4𝑥) + (𝑦² + 6𝑦) = 9
(𝑥² − 4𝑥 + __) + (𝑦² + 6𝑦 + __) = 9 + __ + __
(𝑥² − 4𝑥 + 4) + (𝑦² + 6𝑦 + 9) = 9 + 4 + 9
(𝑥 − 2)² + (𝑦 + 3)² = 22
(𝑥 − 2)² + (𝑦 + 3)² = √22
ℎ = 2, 𝑘 = −3, 𝑟 = √22
Therefore, the center of the circle is at (2, −3) and the radius is √22 units.
Example 5. Write the equation of the circle in center-radius form whose center is at (6,7) and contains the point (2,5).
Solution: Step 1: Find the radius using the distance formula.
𝑑 = √((𝑥₂ − 𝑥₁)² + (𝑦₂ − 𝑦₁)²) Let P₁(6,7) and P₂(2,5)
𝑟 = √((2 − 6)² + (5 − 7)²) Substitute P₁, P₂ and r.
𝑟 = √((−4)² + (−2)²) Perform the operations.
𝑟 = √16 + 4 Square each term.
𝑟 = √20 = 2√5 Simplify.
Step 2: Write the equation in center-radius form.
At 𝐶(6, 7); h = 6, 𝑘 = 7.
(𝑥 − ℎ)² + (𝑦 − 𝑘)² = 𝑟² Use the center-radius form.
(𝑥 − 6)² + (𝑦 − 7)² = (2√5)² Substitute the values of ℎ, 𝑘, and 𝑟.
(𝑥 − 6)² + (𝑦 − 7)² = 20 Simplify.
Therefore, the equation of the circle in center-radius form is (𝒙 − 𝟔)² + (𝒚 − 𝟕)² = 20.
Example 6: Write the equation of the circle in general form with center at (−5, −4) and radius of 3 units.
Solution: Given: h = −5, k = −4, r = 3
[x − (−5)]² + [y − (−4)]² = 3² Substitute given in (𝑥 − ℎ)² + (𝑦 − 𝑘)² = 𝑟².
(𝑥 + 5)² + (𝑦 + 4)² = 9 Simplify.
𝑥² + 10𝑥 + 25 + 𝑦² + 8𝑦 + 16 = 9 Expand by squaring a binomial.
𝑥² + 𝑦² + 10𝑥 + 8𝑦 + 32 = 0 Write in 𝑥² + 𝑦² + 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0.
Therefore, the equation of the circle in general form is 𝒙² + 𝒚² + 𝟏𝟎𝒙 + 𝟖𝒚 + 𝟑𝟐 = 𝟎.
What’s More
Activity 2: Pair them Up
Match the given parts of the circle with its corresponding equation. Use the letters of your answer to decode the message below.
| Item | Given | Equation |
|---|---|---|
| 1 | Center (0,5) r = 6 | N. (x – 2)² + (y – 4)² = 36 |
| 2 | Center (8, −7) r = 15 | R. (x + 4)² + (y – 9)² = 144 |
| 3 | Center (6,1) r = 9 | A. (x – 6)² + (y – 1)² = 81 |
| 4 | Center (−4,9) r = 12 | E. (x – 8)² + (y + 7)² = 225 |
| 5 | Center (2,4) r = 6 | L. x² + (y – 5)² = 36 |
| 6 | Center (−4, −4) r = 4√2 | H. (x – 7)² + y² = 64 |
| 7 | Center (5,5) r = 3√3 | T. x² + (y + 2)² = 49 |
| 8 | Center (−2,0) r = 10 | M. (x + 2)² + y² = 100 |
| 9 | Center (0, −2) r = 7 | G. (x – 5)² + (y – 5)² = 27 |
| 10 | Center (7,0) r = 8 | I. (x + 4)² + (y + 4)² = 32 |
| 11 | Center (−2,2) r = 4√2 | F. x² + y² – 6x – 10y + 18 = 0 |
| 12 | Center (3,0) r = 4 | Y. x² + y² – 6x – 7 = 0 |
| 13 | Center (3,5) r = 4 | S. x² + y² + 4x – 4y – 24 = 0 |
| 14 | Center (0,0) r = 5 | O. (x – 3)² + (y – 3)² = 20 |
| 15 | Center (3,3) r = 2√5 | U. x² + y² = 25 |
[transcriber’s note: the decoding grid is printed as three blocks of boxes, one with the numbers 15 10 over 6 11, one with 1 2 3 4 5 6 5 7 over 4 2 3 1 1 12, and one with 8 3 9 10 over 13 14 5 followed by “!”. The boxes are blank except for the numbers listed, which are transcribed here in reading order.]
15 10 1 2 3 4 5 6 5 7 8 3 9 10
6 11 4 2 3 1 1 12 13 14 5 !
What I Have Learned
Complete the statements below:
I have learned that …
when the center of a circle is at the origin, the equation to be used is _____________________________.
when the center of a circle is at any other point (ℎ, 𝑘), the equation to be used is _____________________________.
a circle can also be represented in its general form which is written as _____________________________.
for a circle to exist, radius is _________________________ and coefficients of 𝑥² and 𝑦² should not be equal to _________________________.
given the coordinates of the center and any other point on the circle, the length of the radius can be computed by using the _________________________.
to write the equation of a circle, the two necessary parts must be obtained: (a) _____________________________ and (b) _____________________________.
What I Can Do
Word Problem:
Mang Cardo tied his carabao to a peg in the field. In the evening, dark clouds began to form and the weather turned bad. Consecutive lightning and thunder frightened the carabao causing it to run in circles straining its rope.
In the morning when the weather calmed down, Mang Cardo sent his Grade 10 son to untie the carabao. His son was surprised to see a perfectly formed circle around the carabao’s peg.
If the circle formed by the carabao’s foot prints is given by the equation 𝑥² + 𝑦² − 6𝑥 + 8𝑦 − 119 = 0, how long is the carabao’s rope if the scale used is 1 meter is to 2 units?
| Illustration: | Solution: |
|---|---|
Conclusion:
Assessment
DIRECTION: Read and understand each item, then choose the letter of your answer and write it on your answer sheet.
What is the center-radius form of a circle defined by 𝑥² + 𝑦² − 2𝑥 + 4𝑦 − 1 = 0?
- A) (𝑥 + 1)² + (𝑦 − 2)² = 6
- C) (𝑥 + 1)² + (𝑦 − 2)² = 8
- B) (𝑥 − 1)² + (𝑦 + 2)² = 6
- D) (𝑥 − 1)² + (𝑦 + 2)² = 8
Which of the following is NOT an equation of a circle in general form?
- A) 2𝑥² + 2𝑦² − 12𝑥 − 20𝑦 + 36 = 0
- C) 𝑥² + 𝑦² + 10𝑦 + 9 = 0
- B) 𝑥² + 𝑦² − 6𝑥 − 7 = 0
- D) 𝑥² + 5𝑥 − 6𝑦 + 6 = 0
What is the location of the center of a circle define by (𝑥 + 6)² + (𝑦 + 4)² = 9?
- A) Quadrant I
- B) Quadrant II
- C) Quadrant III
- D) Quadrant IV
What is the equation of the circle whose center is at (4,0) and the radius is 3?
- A) 𝑥² + 𝑦² − 8𝑥 + 4𝑦 + 7 = 0
- C) 𝑥² + 𝑦² − 8𝑥 + 7 = 0
- B) 𝑥² + 𝑦² + 8𝑥 − 4𝑦 − 7 = 0
- D) 𝑥² + 𝑦² − 8𝑦 + 7 = 0
What is the measure of the radius of a circle defined by 𝑥² + 𝑦² + 8𝑥 + 8 = 0?
- A) 2√2
- B) 4√2
- C) 4
- D) 8
The center of a circle is at (7, −2) and the radius is 5 units. What is the equation of the circle?
- A) (𝑥 + 7)² + (𝑦 − 2)² = 25
- C) (𝑥 + 7)² + (𝑦 − 2)² = 5
- B) (𝑥 − 7)² + (𝑦 + 2)² = 25
- D) (𝑥 − 7)² + (𝑦 + 2)² = 5
For items 7 – 8. Consider the circle defined by 3𝑥² + 3𝑦² − 6𝑥 − 6𝑦 + 3 = 0.
What is the center-radius form of the equation of the circle?
- A) (𝑥 − 1)² + (𝑦 − 1)² = 1
- C) (𝑥 + 1)² + (𝑦 − 1)² = 1
- B) (𝑥 + 1)² + (𝑦 + 1)² = 1
- D) (𝑥 − 1)² + (𝑦 + 1)² = 1
What is the center of the circle?
- A) (−1,1)
- B) (−1, −1)
- C) (1, −1)
- D) (1,1)
What is the equation of a circle whose radius is √13 units and its center is at the origin?
- A) 𝑥² + 𝑦² = √13
- C) (𝑥 + 1)² + (𝑦 + 1)² = 13
- B) 𝑥² + 𝑦² = 13
- D) (𝑥 + 1)² + (𝑦 + 1)² = √13
What is the radius of the circle given by the equation 𝑥² + 𝑦² − 8𝑥 + 2𝑦 − 32 = 0?
- A) 5
- B) 6
- C) 7
- D) 8
Additional Activity
DIRECTION: Match the figure with the corresponding equation that represents it. Write the letter on each item.
[illustration: circle centered at the origin with radius 8 drawn from the center; labels “C(0,0)” and “r=8”]
[illustration: circle centered at (5,0) with a vertical radius of 4 drawn downward from the center; labels “C(5,0)” and “r=4”]
[illustration: circle centered at (4,4) with a horizontal radius of 4 drawn to the right; labels “C(4,4)” and “r=4”]
[illustration: circle with center C(6,7) and a marked point P(2,5) on the lower left side of the circle; labels “C(6,7)” and “P(2,5)”]
[illustration: circle centered at C(−3,0) with point P(2,12) on its top; labels “P(2,12)” and “C(−3,0)”]
A) (x – 6)² + (y – 7)² = 20
B) (x + 3)² + y² = 169
C) x² + y² = 64
D) (x – 4)² + (y – 4)² = 16
E) (x – 5)² + y² = 16
Answer Key
What I Know
- D 2) C 3) A 4) D 5) C 6) B 7) B 8) B [?] 9) C 10) A
[transcriber’s note: the item 8 answer is read here as B; the printed letter is hard to confirm, so it is marked with [?]]
[transcriber’s note: this page is printed upside down; the text is transcribed here in its normal reading orientation]
What’s More (Activity 2)
- L 2) E 3) A 4) R 5) N 6) I 7) G 8) M 9) T 10) H 11) S 12) Y 13) F 14) U 15) N
Message: OH LEARNING MATH IS REALLY FUN!
What I Can Do
The carabao’s rope measures 6 meters.
Assessment
- B 2) D 3) C 4) C 5) A 6) B 7) A 8) D 9) B 10) C
Additional Activity
- C 2) E 3) D 4) A 5) B
References
Callanta, Melvin M., Canonigo, Allan M., Chua, Arnaldo I., Cruz, Jerry D., Esparrago, Mirla S., Garcia, Elino S., Magnaye, Aries N., Orines, Fernando B., Perez, Rowena S., Ternida, Conception S.2015. Mathematics-Grade 10 Learner’s Module, Meralco Avenue, Pasig City: REX Book Store, Inc.
Nivera, Gladys C., and Lapinid, Minie Rose C.,Grade 10 Mathematics Patterns and Practicalities. Chino Roces Avenues, Makati City: Don Bosco Press
Boyd, Cindy J., 1998,Glencoe Geometry-Integration, Application, Connections, Westerville, OH USA, The McGraw-Hill Companies, Inc.
For inquiries or feedback, please write or call:
Department of Education – Bureau of Learning Resources (DepEd-BLR)
Ground Floor, Bonifacio Bldg., DepEd Complex Meralco Avenue, Pasig City, Philippines 1600
Telefax: (632) 8634-1072; 8634-1054; 8631-4985
Email Address: blr.lrqad@deped.gov.ph * blr.lrpd@deped.gov.ph
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